Question

solve power series solution

y'' - (x+1) y' -y = 0

Answer 1

$y=\sum c_nx^n \newline y'=\sum nc_nx^{n-1} \newline y''=\sum n(n-1)c_nx^{n-2} \newline \text{substitute the above value in the given equation, we get}\newline \implies \sum_{n=2}^\infty n(n-1)c_nx^{n-2}-(x+1)\sum_{n=1}^\infty nc_nx^{n-1}-\sum_{n=0}^\infty c_nx^n=0 \newline \implies \sum_{n=0}^\infty (n+2)(n+1)c_{n+2}x^{n}-\sum_{n=1}^\infty nc_{n}x^{n}-\sum_{n=0}^\infty (n+1)c_{n+1}x^{n} -\sum_{n=0}^\infty c_nx^n=0 \newline \implies 2c_2-c_1-c_0+\sum_{n=1}^\infty (n+2)(n+1)c_{n+2}x^{n}-\sum_{n=1}^\infty nc_{n}x^{n}-\sum_{n=1}^\infty (n+1)c_{n+1}x^{n} -\sum_{n=1}^\infty c_nx^n=0 \newline \implies 2c_2-c_1-c_0+\sum_{n=1}^\infty [(n+2)(n+1)c_{n+2}- nc_n-(n+1)c_{n+1}-c_n]x^n=0 \newline \text{We get,}\newline 2c_2-c_1-c_0=0\\ \text{and the recurrence relation is,}\\ (n+2)(n+1)c_{n+2}- nc_n-(n+1)c_{n+1}-c_n=0\\ (n+2)(n+1)c_{n+2}- (n+1)c_{n+1}-(n+1)c_n=0\\ c_{n+2}=\frac{(n+1)c_{n+1}+(n+1)c_n}{ (n+2)(n+1)}\\ c_{n+2}=\frac{c_{n+1}+c_n}{(n+2)}\\ \text{Thus, the solution is given by }\newline y=c_0y1+c_1y_2.\\ Let c_0=1, c_1=0, \implies y=y_1.\\ Then, c_2=1/2, c_3=3/4.\\ Therefore, y_1=1+\frac{1}{2}x^2+\frac{1}{4}x^3+•••.\\ Similarly,\newline c_0=0, c_1=1, \implies y=y_2.\newline Then, c_2=1/2, c_3=3/4.\newline Therefore, y_2=x+\frac{1}{2}x^2+\frac{3}{4}x^3+•••.$